Question

已知数an的递推关系为an+1=2/3an+4,且a1=1，求an的通项公式

Answer 1

a(n+1)+x=2/3an+4+x=2/3[an+(4+x)*3/2]

令x=(4+x)*3/2
x=-12
a(n+1)-12=2/3(an-12)
所以an-12是等比数列，q=2/3
an-12=(a1-12)*q^(n-1)
所以an=12-11*(2/3)^(n-1)

Answer 2

3a(n+1)=2an+12；
3(a(n+1)+x)=2(an+x);
2x-3x-3=12;
x=-15;
3(a(n+1)-15)=2(an-15);
an-15=(an-15)/(a(n-1)-15)*((a(n-1)-15)/(a(n-2)-15))*...*(a2-15)/(a1-15)*(a1-15)
=(2/3)^(n-1)*(-14)
an=-14*(2/3)^(n-1)+15;

Answer 3

a(n+1)=2/3 an + 4,
设[a(n+1)+Z]=2/3 [an+Z],即有a(n+1)=2/3 an- 1/3Z
则-1/3 Z=4，Z=-12（代回上式）
∴{an-12}是以2/3为公比，-11为首项的等比数列
∴an-12=(-11)×(2/3)^(n-1)
∴an=(-11)×(2/3)^(n-1)+12